<p>Evaluate \(\displaystyle\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}\) [JEE Main 2015]</p>
Step-by-Step Solution
Key Concept: \intdx/\sqrt{1-x^2} = arcsin x + C. [arcsin x]_0^(1/2) = arcsin(1/2) - 0 = \pi/6.
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<p>$$\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=[\arcsin x]_0^{1/2}=\arcsin\frac{1}{2}-0=\frac{\pi}{6}$$</p>
Correct Answer: A